The following statement seems like it should be trivial: Consider the stochastic integral of a process $F$ with respect to a continuous local martingale $M$. Denote by $M_t^\epsilon = M_{t + \epsilon}$ the martingale started at time $\epsilon$. Similarly, denote $F_t^\epsilon = F_{t + \epsilon}$. Then $$ \int_0^t F_{s}^\epsilon \, dM_s^\epsilon = \int_0^{t+\epsilon} 1_{[\epsilon, \infty)}(s) \, F_s \, dM_s $$ as stochastic processes in $t$. This is effectively an identity for translating the domain of integration. First, I would like to check whether I have a correct proof of the above statement. More importantly, I would like to know whether there is a more direct way of showing this result than through the Riemann approximation used below.
Let $X_t = \int_0^t F_s^\epsilon \, dM_s^\epsilon$ denote the left-hand side above, and let $Y_t = \int_0^t 1_{[\epsilon, \infty)}(s) \, F_s \, dM_s$ so that the right-hand side is $Y_{t + \epsilon}$. Because the processes $(X_t)_{t \geq 0}$ and $(Y_{t+\epsilon})_{t \geq 0}$ are continuous, being stochastic integrals with respect to $M$, it suffices to show that for every $t \geq 0$, $X_t = Y_{t + \epsilon}$ almost surely. If this is true, then the two processes are indistinguishable. The idea now is to represent $X_t$ and $Y_{t+\epsilon}$ as the same limit in probability of Riemann sums. More precisely, if $\Delta_n$ is a partition of $[0,t+\epsilon]$ with mesh $|\Delta_n|$ and with points $t_i$, one of which is $\epsilon$, then $$ Y_{t+\epsilon} = \lim_{|\Delta_n| \to 0} \sum_{i} 1_{[\epsilon,\infty)}(t_i) F_{t_i} (M_{t_{i+1}} - M_{t_i}) $$ as a limit in probability. If $\Delta_n'$ is the partition of $[0,t]$ consisting of the points $s_i = t_i - \epsilon$ that are nonnegative, then the right-hand side above equals $$ \lim_{|\Delta_n'| \to 0} \sum_i F_{s_i}^\epsilon (M_{s_{i+1}}^\epsilon - M_{s_i}^\epsilon) = X_t ,$$ which shows that $X_t = Y_{t+\epsilon}$ almost surely, and the result follows.